In a uniform magnetic field of $0.049 \text{ T}$, a magnetic needle performs $20$ complete oscillations in $5$ seconds as shown. The moment of inertia of the needle is $9.8 \times 10^{-6} \text{ kg m}^2$. If the magnitude of magnetic moment of the needle is $x \times 10^{-5} \text{ Am}^2$, then the value of '$x$' is:

1. Calculate Time Period ($T$): The needle performs 20 oscillations in 5 seconds. $T = \frac{\text{Total Time}}{\text{Number of Oscillations}} = \frac{5}{20} = \frac{1}{4} \text{ s} = 0.25 \text{ s}$
2. Formula: The time period of oscillation for a magnetic needle in a magnetic field $B$ is given by: $T = 2\pi \sqrt{\frac{I}{mB}}$ Where:

• $I$ is the moment of inertia.
• $m$ is the magnetic moment.
• $B$ is the magnetic field strength.

3. Rearrange for Magnetic Moment ($m$): Squaring both sides: $T^2 = 4\pi^2 \frac{I}{mB}$ $m = \frac{4\pi^2 I}{T^2 B}$
4. Substitute Values:

• $I = 9.8 \times 10^{-6} \text{ kg m}^2$
• $B = 0.049 \text{ T} = 49 \times 10^{-3} \text{ T}$
• $T = 1/4 \text{ s} \implies T^2 = 1/16 \text{ s}^2$

$m = \frac{4\pi^2 \times (9.8 \times 10^{-6})}{(1/16) \times (49 \times 10^{-3})}$ $m = \frac{4\pi^2 \times 9.8 \times 10^{-6} \times 16}{49 \times 10^{-3}}$

Notice that $9.8$ is $2 \times 4.9$, and $49$ is $10 \times 4.9$. Let's simplify: $m = \frac{64\pi^2 \times 9.8 \times 10^{-6}}{49 \times 10^{-3}}$ $\frac{9.8}{49} = \frac{1}{5} = 0.2$ $m = 64\pi^2 \times 0.2 \times 10^{-3}$ $m = 12.8\pi^2 \times 10^{-3}$ $m = 1280\pi^2 \times 10^{-5} \text{ Am}^2$ 5. Find $x$: The problem states $m = x \times 10^{-5} \text{ Am}^2$. Comparing values: $x = 1280\pi^2$.